Analyze groups of order \(pq\) with \(q<p\) prime.
- Show that \(G\) is never simple.
- Show that if \(q\nmid p-1\), then \(G\) is cyclic.
- Classify \(G\) when \(q\mid p-1\).
Solution
Let \(|G|=pq\) with primes \(q<p\).
1. The Sylow \(p\)-subgroup is normal.
Its number \(n_p\) satisfies \[ n_p\mid q,\qquad n_p\equiv1\pmod p. \] Since \(q<p\), the possibility \(n_p=q\) is impossible, hence \(n_p=1\). Thus the Sylow \(p\)-subgroup \(P\cong C_p\) is normal, so \(G\) is not simple.
2. Every such group is a semidirect product \(C_p\rtimes C_q\).
Let \(Q\) be a Sylow \(q\)-subgroup. Then \(P\cap Q=1\) and \(|PQ|=pq\), so \(G=PQ\). Since \(P\trianglelefteq G\), \[ G\cong P\rtimes Q\cong C_p\rtimes_\theta C_q, \] where \(\theta:C_q\to\operatorname{Aut}(C_p)\cong C_{p-1}\).
3. If \(q\nmid p-1\), then \(G\cong C_{pq}\).
The image of \(\theta\) has order dividing both \(q\) and \(p-1\). If \(q\nmid p-1\), the image is trivial. Hence \[ G\cong C_p\times C_q\cong C_{pq}. \]
4. If \(q\mid p-1\), there are exactly two isomorphism types.
Because \(C_{p-1}\) is cyclic, it has a unique subgroup of order \(q\). Thus there is the trivial action, giving \(C_{pq}\), and a nontrivial action with image that unique subgroup. Any two nontrivial homomorphisms \(C_q\to C_{p-1}\) differ by an automorphism of \(C_q\), hence yield isomorphic semidirect products. Therefore the two groups are \[ C_{pq},\qquad C_p\rtimes C_q \] with the latter the unique nonabelian isomorphism type.